A Differential relates the angular positions and forces of three gears on a common axis of rotation. They are commonly used in car steering systems to make sure that when a car rounds a corner, the inside wheel spins less than the outside wheel. In our erg, we use the differential to split an input force from the handle to the two output forces that act on flywheel and the tires.
Subscripts
| Latex |
|---|
\large L, R, M |
- When we hold the right axle fixed and turn the left axle one revolution, the middle sprocket makes half a revolution in the same angular direction.
- Holding the middle sprocket in place and turning the left axle one revolution causes the right axle to make one revolution in the opposite direction.
- The above observations are the same if the left and right axle are switched.
The results can be summarized as such:
| Latex |
|---|
\large\begin{align* |
| Wiki Markup |
!differentialwithangles.png|border=1! {latex} Subscripts {latex} $L, R, M$ {latex} correspond the left half-shaft, right half-shaft, and middle gear, respectively. Three observations: \begin{enumerate} \item When we hold the right axle fixed and turn the left axle one revolution, the middle sprocket makes half a revolution in the same angular direction. \item Holding the middle sprocket in place and turning the left axle one revolution causes the right axle to make one revolution in the opposite direction. \item The above observations are the same if the left and right axle are switched. \end{enumerate} The results can be summarized as such: \begin{equation} \theta_{\text{M}} = \frac{1}{2}\theta_{\text{L}} + \frac{1}{2}\theta_{\text{R}} \end{equationalign*} |
where
| Latex |
|---|
\large $\theta$ |
...
angular
...
position. In other words, the position of the middle gear is half the sum of the left and right gears. Say that
| Latex |
|---|
\large $\omega$ |
...
angular
...
velocity,
| Latex |
|---|
\large $\alpha$ |
...
...
angular
...
acceleration,
...
and
| Latex |
|---|
\large $\tau$ |
...
torque.
| Latex |
|---|
\large $r$ |
...
the
...
radius
...
of
...
the
...
gear,
...
and
| Latex |
|---|
\large $v$ |
...
the
...
linear
...
velocity
...
at
...
the
...
edge
...
of
...
a
...
gear. From the first equation, we can find the relationships between the angular and linear velocities of the gears.
| Latex |
|---|
\large \begin{align*} \omega_{\text{M}} =& \frac{1}{2} \left( \omega_{\text{L}} + \omega_{\text{R}} \right) \\ \omega_{\text{M}} r_{\text{M}} =& \frac{1}{2} ( \omega_{\text{L}} r_{\text{M}} + \omega_{\text{R}} r_{\text{M}}) \\ v_{\text{M}} =& \frac{1}{2} \left( \omega_{\text{L}} r_{\text{M}} \frac{r_{\text{L}}}{r_{\text{L}}} + \omega_{\text{R}} r_{\text{M}} \frac{r_{\text{R}}}{r_{\text{R}}} \right) \\ v_{\text{M}} =& \frac{1}{2} \left( v_{\text{L}} \frac{r_{\text{M}}}{r_{\text{L}}} + v_{\text{R}} \frac{r_{\text{M}}}{r_{\text{R}}} \right) \end{align*} |
We
...
assume
...
that
...
these
...
ideal
...
gears
...
are
...
...
and
...
massless.
...
Therefore,
...
we
...
can
...
use
...
conservation
...
of
...
energy
...
to
...
say
...
that
...
input
...
power
...
equals
...
output
...
power.
...
Say
| Latex |
|---|
\large $P$ |
...
power
...
as
...
a
...
function
...
of
...
time.
| Latex |
|---|
\large \begin{align*} &P_{\text{M}} = P_{\text{L}} + P_{\text{R}} \\ &\tau_{\text{M}} \cdot \omega_{\text{M}} = \tau_{\text{L}} \cdot \omega_{\text{L}} + \tau_{\text{R}} \cdot \omega_{\text{R}} \\ &\tau_{\text{M}} \cdot \left[ \frac{1}{2} \left( \omega_{\text{L}} + \omega_{\text{R}} \right) \right] = \tau_{\text{L}} \cdot \omega_{\text{L}} + \tau_{\text{R}} \cdot \omega_{\text{R}} \end{align*} |
If
...
we
...
consider
| Latex |
|---|
\large $\omega_{\text{L}}$ |
| Latex |
|---|
$\omega_{\text{R}}$ |
...
we
...
find
...
that
| Latex |
|---|
\large \begin{align*} \frac{1}{2} \tau_{\text{M}} \omega_{\text{L}} &= \tau_{\text{L}} \omega_{\text{L}}, \;\;\; \frac{1}{2} \tau_{\text{M}} \omega_{\text{R}} = \tau_{\text{R}} \omega_{\text{R}} \\ \frac{1}{2} \tau_{\text{M}} &= \tau_{\text{L}}, \;\;\;\;\;\;\; \;\; \; \;\; \frac{1}{2} \tau_{\text{M}} = \tau_{\text{R}} \end{align*} |
Therefore,
...
we
...
can
...
conclude
...
that
...
the
...
relationships
...
between
...
torques
...
is
...
the
...
same
...
as
...
the
...
relationships
...
between
...
angular
...
position; a torque on the middle gear is evenly divided between the torque on the left gear and the torque on the right gear.
| Latex |
|---|
\large \begin{align*} \tau_{\text{M}} &= \tau_{\text{L}} + \tau_{\text{R}} \\ \tau_{\text{L}} &= \tau_{\text{R}} = \frac{1}{2} \tau_{\text{M}} \end{align |
...
*}
|
| Widget Connector | ||
|---|---|---|
|